4.4 Exponential and Logarithmic Equations
4.4.1 Learning Objectives
By the end of this section, you should be able to:
- solve exponential equations by rewriting with a common base;
- solve exponential equations using logarithms;
- solve equations of quadratic type;
- solve logarithmic equations; and
- identify and reject extraneous solutions.
4.4.2 Exponential Equations with a Common Base
If an equation can be written as \(a^u=a^v\), where \(a>0\) and \(a\ne1\), then \(u=v\).
Example 4.15 Rewriting with a Common Base
Solve \(8^{x-1}=4^{2x+1}\).
Solution
\[ 2^{3x-3}=2^{4x+2}. \]
Equate exponents:
\[ 3x-3=4x+2, \]
so
\[ \boxed{x=-5}. \]
Exercise 4.15 Solve \(9^{x+1}=27^{x-1}\).
Show answer
Answer: \(3^{2x+2}=3^{3x-3}\), so \(\boxed{x=5}\).
4.4.3 Solving Exponential Equations with Logarithms
When common bases are not convenient, isolate the exponential expression, take a logarithm of both sides, use the power law, and solve.
Example 4.16 Using Natural Logarithms
Solve \(5^{2x-1}=17\).
Solution
\[ (2x-1)\ln5=\ln17. \]
Therefore,
\[ 2x-1=\frac{\ln17}{\ln5} \]
and
\[ \boxed{x=\frac12\left(1+\frac{\ln17}{\ln5}\right)\approx1.380}. \]
Exercise 4.16 Solve \(3e^{0.4x}=20\) to three decimal places.
Show answer
Answer: \(e^{0.4x}=20/3\), so \(\boxed{x=\ln(20/3)/0.4\approx4.743}\).
4.4.4 Equations of Quadratic Type
An equation containing \(a^{2x}\) and \(a^x\) can often be treated as a quadratic by substituting \(u=a^x\). Since \(a^x>0\), reject any non-positive value of \(u\).
Example 4.17 Using an Exponential Substitution
Solve \(4^x-5(2^x)+4=0\).
Solution
Since \(4^x=(2^x)^2\), let \(u=2^x\):
\[ u^2-5u+4=0. \]
Thus, \(u=1\) or \(u=4\). Therefore,
\[ 2^x=1\quad\text{or}\quad2^x=4, \]
giving
\[ \boxed{x=0\text{ or }x=2}. \]
Exercise 4.17 Solve \(9^x-10(3^x)+9=0\).
Show answer
Answer: Let \(u=3^x\). Then \((u-1)(u-9)=0\), so \(\boxed{x=0\text{ or }x=2}\).
4.4.5 Logarithmic Equations
For a single logarithm, convert to exponential form. For several logarithms with the same base, combine them first. Every proposed solution must make each original logarithm’s argument positive.
Example 4.18 Solving and Checking a Logarithmic Equation
Solve
\[ \log_2(x-1)+\log_2(x-3)=3. \]
Solution
The domain requires \(x>3\). Combine:
\[ \log_2[(x-1)(x-3)]=3. \]
Convert to exponential form:
\[ (x-1)(x-3)=8. \]
\[ x^2-4x-5=0, \]
so \(x=5\) or \(x=-1\). Only \(x=5\) lies in the domain.
\[ \boxed{x=5} \]
Exercise 4.18 Solve \(\ln(x+4)-\ln x=\ln3\).
Show answer
Answer: The domain is \(x>0\). Since \((x+4)/x=3\), \(x+4=3x\), so \(\boxed{x=2}\).
4.4.6 Graphical Solutions
Desmos can solve equations by graphing each side as a separate function and locating intersections. This is useful when an exact algebraic solution is unavailable. Report only intersections in the original domain.
Example 4.19 Solving Graphically
Use Desmos to solve \(2^x=x+2\).
Solution
Graph \(y=2^x\) and \(y=x+2\). The graphs intersect at \(x=-1\) and \(x=2\).
\[ \boxed{x=-1\text{ or }x=2} \]
Exercise 4.19 Use Desmos to solve \(e^x=5-x\) to two decimal places.
Show answer
Answer: \(\boxed{x\approx1.31}\)
4.4.7 Conceptual Takeaways
- Equal exponential expressions with the same base have equal exponents.
- Logarithms bring an exponent down as a coefficient.
- Exponential substitution can turn some equations into quadratics.
- Logarithmic equations require positive arguments.
- Graphical solutions are approximations unless the intersections are known exactly.
4.4.8 Skills You Should Be Able to Do
- Solve exponential equations using common bases.
- Apply logarithms to isolated exponential expressions.
- Solve exponential equations of quadratic type.
- Combine and solve logarithmic equations.
- Check domains and verify solutions graphically.
4.4.9 Practice Problems with Solutions
Solve \(2^{x+3}=32\).
Show Solution
Since \(32=2^5\), \(x+3=5\). Thus, \(\boxed{x=2}\).
Solve \(5^{2x}=125\).
Show Solution
Since \(125=5^3\), \(2x=3\), so \(\boxed{x=3/2}\).
Solve \(7^x=20\) to three decimal places.
Show Solution
\[ x\ln7=\ln20, \]
so
\[ \boxed{x=\frac{\ln20}{\ln7}\approx1.539}. \]
Solve \(e^{3x-1}=9\).
Show Solution
\[ 3x-1=\ln9, \]
so
\[ \boxed{x=\frac{1+\ln9}{3}\approx1.066}. \]
Solve \(4^x-6(2^x)+8=0\).
Show Solution
Let \(u=2^x\). Then
\[ u^2-6u+8=(u-2)(u-4)=0. \]
Hence, \(2^x=2\) or \(2^x=4\), giving \(\boxed{x=1\text{ or }x=2}\).
Solve \(\log_3(x+1)=2\).
Show Solution
Convert to exponential form:
\[ x+1=3^2=9. \]
Therefore, \(\boxed{x=8}\).
Solve \(\ln x+\ln(x-2)=\ln8\).
Show Solution
The domain requires \(x>2\). Combine:
\[ x(x-2)=8. \]
Then \(x^2-2x-8=0\), so \(x=4\) or \(x=-2\). Reject \(-2\).
\[ \boxed{x=4} \]